帮我写一个单片机八路抢答器
时间: 2023-08-20 11:25:27 浏览: 43
好的,下面是一个简单的单片机八路抢答器的代码实现,使用的是STC89C52单片机:
```c
#include <reg52.h>
// 定义按键和LED灯的IO口
sbit button0 = P2^0;
sbit button1 = P2^1;
sbit button2 = P2^2;
sbit button3 = P2^3;
sbit button4 = P2^4;
sbit button5 = P2^5;
sbit button6 = P2^6;
sbit button7 = P2^7;
sbit led0 = P1^0;
sbit led1 = P1^1;
sbit led2 = P1^2;
sbit led3 = P1^3;
sbit led4 = P1^4;
sbit led5 = P1^5;
sbit led6 = P1^6;
sbit led7 = P1^7;
// 定义按键状态数组和LED灯状态数组
unsigned char button_state[8] = {1, 1, 1, 1, 1, 1, 1, 1};
unsigned char led_state[8] = {0, 0, 0, 0, 0, 0, 0, 0};
// 延时函数
void delay(unsigned int i) {
while(i--);
}
// 检测按键状态
void check_button() {
if (button0 == 0) {
button_state[0] = 0;
}
if (button1 == 0) {
button_state[1] = 0;
}
if (button2 == 0) {
button_state[2] = 0;
}
if (button3 == 0) {
button_state[3] = 0;
}
if (button4 == 0) {
button_state[4] = 0;
}
if (button5 == 0) {
button_state[5] = 0;
}
if (button6 == 0) {
button_state[6] = 0;
}
if (button7 == 0) {
button_state[7] = 0;
}
}
// 更新LED灯状态
void update_led() {
if (button_state[0] == 0 && led_state[0] == 0) {
led_state[0] = 1;
led0 = 1;
}
if (button_state[1] == 0 && led_state[1] == 0) {
led_state[1] = 1;
led1 = 1;
}
if (button_state[2] == 0 && led_state[2] == 0) {
led_state[2] = 1;
led2 = 1;
}
if (button_state[3] == 0 && led_state[3] == 0) {
led_state[3] = 1;
led3 = 1;
}
if (button_state[4] == 0 && led_state[4] == 0) {
led_state[4] = 1;
led4 = 1;
}
if (button_state[5] == 0 && led_state[5] == 0) {
led_state[5] = 1;
led5 = 1;
}
if (button_state[6] == 0 && led_state[6] == 0) {
led_state[6] = 1;
led6 = 1;
}
if (button_state[7] == 0 && led_state[7] == 0) {
led_state[7] = 1;
led7 = 1;
}
}
// 主函数
void main() {
// 初始化按键状态和LED灯状态
unsigned char i;
for (i = 0; i < 8; i++) {
button_state[i] = 1;
led_state[i] = 0;
}
// 循环检测按键状态和更新LED灯状态
while (1) {
check_button();
update_led();
delay(1000);
}
}
```
这个代码实现了一个简单的单片机八路抢答器,当有一个按钮被按下时,对应的LED灯会亮起表示该按钮被抢答成功。希望能对你有所帮助!